Calculate enthalpy of the reaction,
`F_(2)O(g)+H_(2)O(g) to O_(2)(g)+2HF(g)`
if the bond energies of O-F,O-H,H-F and O=O bonds are 44, 111, 135 and 119 kcal per mole respectively.

Explanation

We have to calculate `DeltaH` of the reaction
`F_(2)O(g)+H_(2)O(g) to O_(2)(G)+2HF (g) `
For reactants
Bond energy of 2 moles of O-F bonds `=2xx44` kcal
Bond energy of 2 moles of O-H bonds `=2xx111` kcal
For products
Bond formation energy of 1 mole of O=O bonds `=-119` kcal
Bond formation energy of 2 moles of H-F bonds `=-2xx135` kcal
Adding, we get `DeltaH` of the required equation,
`DeltaH=2xx44+2xx111-119-2xx135=-79` kcal